The Square Of Mass Formula And Matter Waves

[Dubbelosix]

In natural units,the gravitational charge is equal to the square of the mass of a particle

[math]\alpha_G = m^2[/math]

For Matti, the quantization of the mass has a spectral property:

[math]n\hbar = m^2[/math]

[math]\mathbf{R} = \frac{\alpha_G}{4 \pi \lambda_0} = \frac{1}{\hbar c}\frac{Gm^2}{4 \pi \lambda_0} = \frac{1}{\hbar c} \frac{Gm^3c}{4 \pi \hbar}[/math]

This is analogous to the Rydberg constant - which even though was first applied to hydrogen atoms, it could be derived from fundamental concepts (according to Bohr).

In which case we may hypothesize energy levels:

[math]\frac{1}{\lambda} = \mathbf{R}(\frac{1}{n^2_1} - \frac{1}{n^2_2})[/math]

Plugging in the last expressions we get an energy equation:

[math]\Delta E_G = \frac{\hbar c}{\lambda_0} = \frac{1}{4 \pi }\frac{m_0v^2}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{4 \pi \lambda_0}(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2}) = \frac{p}{4 \pi \hbar}(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2})[/math]

The relativistic gamma appears from the definition of the deBroglie wavelength [1] and it suggests a relationship between Einstein's relativistic mass and the gravitational charge.It also implies that motion generates the gravitational charge. Since particles are always in motion, this is a theoretical possibility. The motion generating the gravitational charge could be seen in a similar context to how an electron possesses an electric charge due to motion in an electromagnetic field.In the last term we end up with the deBroglie wavelength, but it is smaller by a factor [math]4 \pi[/math]. It is considered that the deBroglie wavelength is a fundamental concept of all matter. The difference of two energy states depends on the difference of the gravitational charges, weighted by the relativistic wavelength.

 

[1] - The deBroglie relationship used was:

 

[math]\frac{1}{\lambda} = \frac{\gamma m_0 v}{\hbar} = \frac{1}{\hbar} \frac{m_0v}{(1 - \frac{v^2}{c^2})}[/math]

Edited March 5, 2018 by Dubbelosix